Math Help - find equations of tan lines parallel to line

1. find equations of tan lines parallel to line

I am stuck on this problem:

Find the equations of the tangent lines to the graph of the equation below that are parallel to the line given.

f(x) = (x+1)/(x-1)
Line : 2y+x=6

2. Originally Posted by sydewayzlocc
I am stuck on this problem:

Find the equations of the tangent lines to the graph of the equation below that are parallel to the line given.

f(x) = (x+1)/(x-1)
Line : 2y+x=6
1. find the slope, $m$ , of the given line.

2. set $f'(x) = m$ and solve for values of x on the curve.

3. evaluate the corresponding y-value for each x-value found in step 2 ... then use the point-slope form of a linear equation to write the tangent line equation.

3. i am still stuck on this problem, can anyone give me some help?

4. Originally Posted by sydewayzlocc
i am still stuck on this problem, can anyone give me some help?
where are you stuck? ... show any work you have completed.

5. Originally Posted by sydewayzlocc
i am still stuck on this problem, can anyone give me some help?
What is the slope of the line 2y+x=6?

What is the derivative of the function f(x) = (x+1)/(x-1)?

If you have been asked to do this problem then clearly you are expected to know how to do those. Please show what you have done on those.

6. Re: find equations of tan lines parallel to line

I'm stuck on the same problem, m = -1/2

f '(x) = (-3)/(x-1)^2

I set them equal and got x = 5
y = -1/2x + 3/2
but its wrong.

7. Re: find equations of tan lines parallel to line

$f'(x) = -\dfrac{2}{(x-1)^2}$

Hence, your numerator was incorrect. Using the correct derivative, you should be able to get the correct answers (there will be more than one).

$-\dfrac{2}{(x-1)^2} = -\dfrac{1}{2}$

$(x-1)^2 = 4$

$x-1 = \pm 2$

$x = 1 \pm 2$, so either $x = -1$ or $x = 3$.

Then, $f(-1) = 0$ and $f(3) = 2$, so both lines should have slope $-\dfrac{1}{2}$, one should pass through the point $(-1,0)$ and the other through the point $(3,2)$.